second order taylor polynomial

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November 4th 2007, 10:03 AM
dopi

second order taylor polynomial

Hi i was wondering if anyone could help me with this question.

i need to use the second order taylor polynomial with remainder to estimate sqraure-root of 101 and give an estimate of the error mode.

but where i get confused is that i am using the formula for the taylor poly..and i dont know what f(a) so i cant dfferentiate and so i cant fill in the taylor poly formula..all i am is given is square root of 101..

can someone please help thanks
November 4th 2007, 10:41 AM
ThePerfectHacker

Quote:

Originally Posted by dopi

Hi i was wondering if anyone could help me with this question.

i need to use the second order taylor polynomial with remainder to estimate sqraure-root of 101 and give an estimate of the error mode.

but where i get confused is that i am using the formula for the taylor poly..and i dont know what f(a) so i cant dfferentiate and so i cant fill in the taylor poly formula..all i am is given is square root of 101..

can someone please help thanks

Let $f(x) = \sqrt{x+100} = (x+100)^{1/2}$ the Taylor polynomial centered at zero or order 2 is, $T_2(x) = 10 + \frac{x}{20} - \frac{x^2}{4000}$ consider the interval $(-2,2)$ the derivative is $f'(x) = \frac{1}{2}(x+100)^{-1/2}$ which is bounded by $\frac{1}{2}(98)^{1/2}$ .
So the maximum error possible,
$\frac{\sqrt{98}}{2\cdot 3!}$
March 26th 2009, 08:11 AM
varkoume.com

Hello I have the same problem and I cannot see how you estimated the square root (101)
March 27th 2009, 03:52 AM
mr fantastic

Quote:

Originally Posted by varkoume.com

Hello I have the same problem and I cannot see how you estimated the square root (101)

The exact value is $f(1)$ and the estimate is $T_2 (1)$ .